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sergeinik [125]
3 years ago
12

Which statement is an example of the Addition Property of Equality? (1 point). . If p = q then p • s = q • s .. If p = q then p

+ s = q + s .. If p = q then p - s = q - s .. p = q
Mathematics
2 answers:
atroni [7]3 years ago
4 0
The formal name for the property of equality that allows one to add the same quantity to both sides of an equation. This is one of the most commonly used properties for solving equations.

The formula tells us that if a = b, then a + c = b + c. The letters a and b stand for two separate numbers, our two twins, and the letter c stands for what we give to each twin to keep them matching. So, for example, if we add 1 to the left side of the equation, we must also add 1 to the right side of the equation. By doing this, we keep the equation, the twins, the same.

Answer: correct choice is B.
Rama09 [41]3 years ago
3 0
The answer to the question above is the second choice where s is added to both sides of the equation giving the final answer which is, p + s = q + s. All the other choices aside from that chosen do not represent an addition property of equality. 
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Given that tan θ ≈ −0.087, where 3 2 π < θ < 2 , π find the values of sin θ and cos θ.
elena-s [515]

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Step-by-step explanation:

Straightforward use of the inverse tangent function of a calculator will tell you θ ≈ -0.08678 radians. This is an angle in the 4th quadrant, where your restriction on θ places it. (To comply with the restriction, you would need to consider the angle value to be 2π-0.08678 radians. The trig values for this angle are the same as the trig values for -0.08678 radians.)

Likewise, straightforward use of the calculator to find the other function values gives ...

  sin(-0.08678 radians) ≈ -0.08667

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_____

<em>Note on inverse tangent</em>

Depending on the mode setting of your calculator, the arctan or tan⁻¹ function may give you a value in degrees, not radians. That doesn't matter for this problem. sin(arctan(-0.087)) is the same whether the angle is degrees or radians, as long as you don't change the mode in the middle of the computation.

We have shown radians in the above answer because the restriction on the angle is written in terms of radians.

_____

<em>Alternate solution</em>

The relationship between tan and sin and cos in the 4th quadrant is ...

  \cos{\theta}=\dfrac{1}{\sqrt{1+\tan^2{\theta}}}\\\\\sin{\theta}=\dfrac{\tan{\theta}}{\sqrt{1+\tan^2{\theta}}}

That is, the cosine is positive, and the sign of the sine matches that of the tangent.

This more complicated computation gives the same result as above.

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